40 = () + () = () + () = () + () write prime

40 = () + () = () + () = () + () write prime

11+29=3+37=17+23

40 = () + () prime numbers in brackets (numbers cannot be repeated)

3+37=40

Given that A1 = 1 and point (an, an + 1) in the sequence an, let BN satisfy BN = 2 ^ an-1 on the image of function f (x) = x + 2, and find the general term formula of BN
And the first n terms and formula SN

A (n + 1) - an = 2 is a constant
The sequence an is an arithmetic sequence
a1=1,an=2n-1
bn=2^an-1=2^(2n-1)-1
sn=2(1-4^n)/(1-4)-n=2/3×4^n-2/3-n

Let the density function of random variable X be f (x) = ax & # 178;, 0

I don't know if it's right

The range of y = Sin & # 178; X + 4cosx + 1

y=sin²x+4cosx+1
=1-cos²x+4cosx+1
=-cos²x+4cosx+2
=-(cosx-2)^2+6
Cosx belongs to [- 1,1]
When cosx = - 1, there is a minimum value of - 3
When cosx = 1, there is a maximum of 5
So the range is [- 3,5]

There is no function that is both odd and even,

No! F (x) = 0 is both odd and even!

Under what circumstances and how to make the "arc on circle" appear in the "construction" of the Geometer's Sketchpad?

Although an arc on a circle is an arc between two points, its radian is determined by the center of the circle. Therefore, the premise of constructing an arc is to select only the center of the circle, a point on the circle, and another point on the circle. The center of the circle must be selected for the first time, and the two points on the circle determine the direction of the arc

As shown in the figure, given the square ABCD with side length a, e is the midpoint of AD, P is the midpoint of CE, and F is the midpoint of BP, calculate the area of △ BFD

Connect DP, make PM ⊥ CD, PN ⊥ BC, ∵ square ABCD with side length a, e is the midpoint of AD, P is the midpoint of CE, ∵ BC = CD = a, PM = 12ed = 14a, PN = 12a, ∵ s △ BDP = s △ bdc-s △ bpc-s △ DPC = 12a2-12 × a × 12a-12 × a × 14a = 18A2, ∵ f is the midpoint of BP, and the distance from P to BD is f

Given the quadratic function f (x) = AX2 + BX + C, f (2) = 0, f (- 5) = 0, f (0) = 1, find the quadratic function

∵ y = AX2 + BX + C satisfies f (2) = 0, f (- 5) = 0, f (0) = 1, that is, through three points a (2, 0), B (- 5, 0), C (0, 1), ∵ 4A + 2B + C = 025a − 5B + C = 0C = 1, the solution is a = − 110b = − 310c = 1, therefore, the analytic expression of the quadratic function is f (x) = − 110x2 − 310x + 1

There are three piles of chessmen, each pile has the same number of chessmen, and there are only black and white chessmen. The first pile has the same number of black chessmen as the second pile, and the third pile has 1 / 4 of all the black chessmen. Put these three piles of chessmen together, how many parts of the total number of white chessmen account for? No equation! Write a clear formula!

Let the number of pieces in each pile be "1" because three black = 1 / 4 all black, so one black + two black = 3 / 4 all black, because one black = two white, so two black + two white = 3 / 4 all black = second pile = "1" because three black = 1 / (3 / 4) = 4 / 3, so three black = (4 / 3) * (1 / 4) = 1 / 3, three white = 2 / 3, one black + two black = second pile = "1